Cauchy sequence proof.

1. The problem statement, all variables and given/known data
Assume $x_n$ and $y_n$ are Cauchy sequences.
Give a direct argument that $x_n+y_n$ is Cauchy.
That does not use the Cauchy criterion or the algebraic limit theorem.
A sequence is Cauchy if for every $\epsilon>0$ there exists an
$N\in \mathbb{N}$ such that whenever $m,n\geq N$
it follows that $|a_n-a_m|< \epsilon$
3. The attempt at a solution
Lets call $x_n+y_n=c_n$
now we want to show that $|c_m-c_n|< \epsilon$
Lets assume for the sake of contradiction that
$c_m-c_n> \epsilon$
so we would have
$|x_m+y_m-x_n-y_n|> \epsilon$
$x_m> \epsilon+y_n-y_m$
since $y_n>y_m$
and we know that $x_m< \epsilon$
so this is a contradiction and the original statement must be true.

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 Quote by cragar 1. The problem statement, all variables and given/known data Assume $x_n$ and $y_n$ are Cauchy sequences. Give a direct argument that $x_n+y_n$ is Cauchy. That does not use the Cauchy criterion or the algebraic limit theorem. A sequence is Cauchy if for every $\epsilon>0$ there exists an $N\in \mathbb{N}$ such that whenever $m,n\geq N$ it follows that $|a_n-a_m|< \epsilon$ 3. The attempt at a solution Lets call $x_n+y_n=c_n$ now we want to show that $|c_m-c_n|< \epsilon$
It's always good to be careful with wording. We want to show that if n and m are big enough, that this inequality holds.

 since $y_n>y_m$
This is probably not true, especially since you haven't even said what n and m are besides arbitrary numbers!

 and we know that $x_m< \epsilon$
This is also probably not true since there's no reason to think the limit is zero

To get the contradiction you're going to want to use the triangle inequality on |(xn-xm)+(yn-ym)|

 ok thanks for your response. So I take $|(x_n-x_m)+(y_n-y_m)| \leq |x_n-x_m|+|y_n-y_m|$ lets assume that $|x_n-x_m|+|y_n-y_m| > \epsilon$ Im going to rewrite it as $A+B> \epsilon$ so now we have $A> \epsilon -B$ Can I just say this since we know that $A< \epsilon$ and $B< \epsilon$ since $\epsilon$ can be any number bigger than zero, then both of these values should be less than $\frac{\epsilon}{2}$ therefore $A+B< \epsilon$ I have a feeling my last step is not ok

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Cauchy sequence proof.

 Quote by cragar Can I just say this since we know that $A< \epsilon$ and $B< \epsilon$ since $\epsilon$ can be any number bigger than zero, then both of these values should be less than $\frac{\epsilon}{2}$ therefore $A+B< \epsilon$
This is the crux of the argument. It's not the whole proof of course - A and B aren't always that small. Feel free to post a full proof if you want it checked over for errors